Question

If the boiling point of benzene (MW = 78 kg/kmol) at 1 atm pressure is 353 K and the
latent heat of vaporization of liquid benzene at its normal boiling point is 393 kJ/kg,
estimate the vapor pressure of benzene at 298 K.

Answer 1

Answer:

Explanation:

C  

6

​

H  

6

​

 

(l)

​

⟶C  

6

​

H  

6

​

 

(g)

​

 

∴Δn=n  

P

​

−n  

R

​

=1−0=1

Given that:-

ΔH=30.84KJ/mol=30840J/mol

T=353K

As we know that,

ΔH=ΔE+Δn  

g

​

RT

30840=ΔE+1×8.314×353

ΔE=30840−2934.842=27905.158=27.9KJ

Now  

Molecular weight of benzene =78g/mol

Weight of benzene =7.8g

No. of moles of benzene =  

78

7.8

​

=0.1 mole

∴ Energy provided to evaporate 0.1 mole of benzene (E)=30.84×0.1=3084J

As we know that,  

E=V×I×t

Given that,

V=12 volt

t=?

I=0.5A

3084=12×0.5×t

⇒t=  

6

3084

​

=514s

Hence the molar internal energy change will be 27.9KJ and for t=514s, a 12 volt source need to supply a 0.5A current in order to vaporise 7.8g of the sample at its boiling point.